Exercise
3.2
1.
Given that x ≥ y. Fill in the blanks with suitable inequality signs.
Answer:
(i) y ≤ x
(ii) x + 6 ≥ y + 6
(iii) x²≥ xy
(iv) –xy ≤ – y²
(v) ) x – y ≥ 0
2.
Say True or False.
(i)
Linear inequation has almost one solution.
(ii)
When x is an integer, the solutions for x ≤ 0 are −1, −2,...
(iii)
An inequation, −3 < x < −1, where x is an integer, cannot be
represented in the number line.
(iv)
x < −y can be rewritten as −y < x
3.
Solve the following inequations.
(i)
x ≤ 7, where x is a natural number.
(ii)
x − 6 < 1, where x is a natural number.
(iii)
2a + 3 ≤ 13, where a is a whole number.
(iv)
6x − 7 ≥ 35, where x is an integer.
(v)
4x − 9 > −33, where x is a negative integer.
Solution:
(i) x ≤ 7, where x is a
natural number.
Since the solution belongs to the set of natural numbers, that
are less than or equal to 7, we take the values of x as 1,2, 3, 4, 5, 6 and 7.
(ii) x – 6 < 1, where x is
a natural number.
x – 6 < 1 Adding 6 on the both the sides x – 6 + 6 < 1 + 6
x < 7
Since the solutions belongs to the set of natural numbers that
are less than 7, we take the values of x
as 1,2, 3, 4, 5 and 6
(iii) 2a + 3 ≤ 13, where a is a whole number.
2a + 3 ≤ 13
Subtracting 3 from both the sides
2a + 3 – 3 ≤ 13 – 3
2a ≤ 10
Dividing both the sides by 2.
2a/2 ≤ 10/2
a ≤ 5
Since the solutions belongs to the set of whole numbers that are
less than or equal to 5 we take the values of a as 0, 1, 2, 3, 4 and 5
(iv) 6x – 7 ≥ 35, where x is an integer.
6x – 7 ≥ 35 Adding 7 on both the sides
6x –7 + 7 ≥ 35 + 7
6x ≥ 42
Dividing both the sides by 6 we get
6x/6 ≥ 42/6
x ≥ 7
Since the solution belongs to the set of integers that are
greater than or equal to 7, we take the values of x as 7, 8, 9, 10...
(v) 4x – 9 > –33, where x is a negative integer.
4x – 9 > – 33 Adding
9 both the sides
4x – 9 + 9 > –33 +
9
4x > – 24
Dividing both the sides by 4
4x/4 > –24/4
x > – 6
Since the solution belongs to a negative integer that are
greater than –6, we take values of u
as –5, –4, –3, –2 and–1.
4.
Solve the following inequations and represent the solution on the number line:
(i)
k > −5, k is an integer.
Since the solution belongs
to the set of integers, the solution is –4, –3, –2, –1, 0,... It’s graph on number
line is shown below.
(ii)
−7 ≤ y, y is a negative integer.
–7 ≤ y
Since the solution set belongs to the set of negative integers,
the solution is –7,–6,–5,–4,–3,–2,–1.
Its graph on the number line is shown below
(iii)
−4 ≤ x ≤ 8, x is a natural number
–4 ≤ x ≤ 8
Since the solution belongs to the set of natural numbers, the
solution is 1,2, 3,4, 5, 6, 7 and 8.
Its graph on number line is shown below
(iv)
3m − 5 ≤ 2m + 1, m is an integer.
3m – 5 ≤ 2m + 1
Subtracting 1 on both the sides
3m – 5 – 1 ≤ 2m + 1 + 1
3m – 6 ≤ 2m
Subtracting 2m on both the sides
3m – 6 – 2m ≤ 2m
– 2m
m – 6 ≤ 0
Adding 6 on both the sides
m – 6 + 6 ≤ 0 + 6
m ≤ 6
Since the solution belongs to the set of integers, the solution
is 6, 5, 4, 3, 2, 1, 0, –1,...
Its graph on number line is shown below
5.
An artist can spend any amount between ₹ 80 to ₹ 200
on brushes. If cost of each brush is ₹ 5 and there are 6 brushes in each packet,
then how many packets of brush can the artist buy?
Solution:
Given the artist can spend any amount between ₹ 80 to ₹ 200
Let the number of packets of brush he can buy be x
Given cost of 1 brush = ₹ 5
Cost of 1 packet brush (6 brushes) = ₹5 x 6 = ₹30
∴ Cost of x packets of
brushes = 30x
∴ The inequation becomes 80 ≤ 30x ≤ 200
Dividing throughout by 30 we get 80/30 ≤ 30x/30 ≤ 200/30
8/3 ≤ x ≤
20/3 ; 2 (2/3) ≤ x ≤ 6 (2/3)
brush packets cannot get in fractions.
∴ The artist can buy 3 ≤ x
≤ 6 packets of brushes.
or x = 3, 4, 5 and 6
packets of brushes.
Objective Type Questions
6.
The solutions of the inequation 3 ≤ p ≤ 6 are (where p is a natural
number)
(i) 4, 5 and 6
(ii) 3, 4 and 5
(iii) 4 and 5
(iv) 3, 4, 5 and 6
[Answer: (iv) 3,4,5 and 6]
7.
The solution of the inequation 5x + 5 ≤ 15 are (where x is a natural
number)
(i) 1 and 2
(ii) 0, 1 and 2
(iii) 2, 1, 0, −1, −2..
(iv) 1, 2, 3..
[Answer: (i) 1 and
2]
Solution: 5x + 5 ≤ 15
5x < 15 – 5 = 10
x ≤ 10/5 = 2
8.
The cost of one pen is ₹ 8 and it is available in a sealed pack
of 10 pens. If Swetha has only ₹ 500, how many packs of pens can she buy
at the maximum?
(i) 10
(ii) 5
(iii) 6
(iv) 8
[Answer: (iii) 6]
Solution: Price of 1 pen = ₹8
Price of 1 pack = 10 × 8 = 80
Number of packs Swetha can buy = x
80x ≤ 500
8x ≤ 50
x ≤ = 50/8 = 6.25
x is a natural number x
= 1, 2, 3, 4, 5, 6
9.
The inequation that is represented on the number line as shown below is _______
(i) -4 < x < 0
(ii) −4 ≤ x ≤ 0
(iii) −4 < x ≤ 0
(iv) −4 ≤ x < 0
[Answer: * –4 ≤ x
≤ 2]
ANSWERS:
Exercise 3.2
1. (i) y ≤
x (ii) x + 6 ≥ y + 6 (iii) x2 ≥ xy (iv)
–xy≤–y2 (v) x–y ≥ 0
2. (i) False (ii)
False (iii) True (iv) False
3. (i) x = 1, 2, 3, 4, 5, 6 and 7
(ii) x = 1, 2, 3, 4, 5 and 6
(iii) a = 0, 1, 2, 3, 4 and 5
(iv) x =
7, 8, 9, 10, ....
(v) x =
–5, –4, –3, –2 and –1
4. (i) k = – 4, –3, –2, –1, 0, 1, 2, ...
(ii) y = – 7, –6, –5, –4, –3, –2 and –1
(iii) x = 1, 2, 3, 4, 5, 6, 7 and 8
(iv) m = ...–3, -2, –1, 0, 1, 2, 3, 4, 5, 6
5. The artist can buy
3 ≤ x ≤ 6 brushes or x = 3, 4, 5 and 6 brushes.
Objective type questions
6. (iv) 3, 4, 5 and 6
7. (i) 1 and 2
8. (iii) 6
9. (ii) –4 ≤ x ≤ 0 / –4 ≤ x ≤ 2
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.